Question

What point in the feasible region maximizes the objective function?

Answer 1

Did you turn around the first contstraint so it looks rational and then draw the graphs of each line?

Answer 2

Not entirely sure how to make it into a rational

Answer 3

Sorry, I thought "rational" was a poor choice of words. I meant to say it's written unnaturally with the 'y' on the right-hand side. Just turn it around so it looks like youwould expect it to look.

The graph them.

Answer 4

But isn't it done alread in the thrid row? Or am I combining everything into a single equation? (I'm new to this so I'm a little confused)

Answer 5

We have x >= 0 and y >= 0. We need to stay in or bordering the 1st Quadrant.

We have y <= x/3 + 1

y-intercept is (0,1) This could be an important point. Remember it.
x-intercept is (-3,0) That's not around the 1st Quadrant. Ignore it.

We have y <= -x + 3

y-intercept is (0,3) That's farther out than (0,1), so we don't care.
x-intercept is (3,0) This could be an important point. Remember it.

Okay, there is only one more thing to do. Let's take stock and see if we're still makign sense.

Answer 6

Yeah I understand it so far..

Answer 7

There's only one more possible vertex for the feasible region. This must be where the two more complicated constraints meet.

y <= x/3 + 1
y <= -x + 3

Let's see where their edges meet!

x/3 + 1 = -x + 3
x + 3 = -3x + 9
4x = 6
x = 2/3

Substituting back

y = -(2/3) + 3 = -2/3 + 9/3 = 7/3

And we have (2/3,7/3). This is in the 1st Quadrant, so we must consider it.

That's it. We found three points that we need to check out. See which one maximizes the objective function.

Answer 8

2/3 and 7/3 points

Answer 9

That's a single point, x = 2/3 and y = 7/3. It is a little unusual to have a constraint with a positive slope.

The three points to check are: (0,1),(3,0), and (2/3,7/3)

Answer 10

I understand how to substitute in the fractions but the other point is a little confusing.